Solving $\csc(x)+\cot(x) = p$

Question

I have come across a problem in which I must solve an equation of the form:

$$ \csc(y)\ +\ \cot(y)\ = \ p $$ Where $y$ is to be solved in terms of $p$. I realise of course that there are many different solutions due to periodicity.

Asking wolfram provides an answer $\left(y=2\left(\pi n+\tan^{-1}\left(\frac{1}{p}\right)\right)\right)$ but not an understanding, and since I am a beginner in this sort of problem and would appreciate some guidance on how to solve problems like this involving various trigonometric functions.

Answer 1

$$\csc(y)+\cot(y)=\frac1{\sin\theta}+\frac{\cos\theta}{\sin\theta}=\frac{1+\cos\theta}{\sin\theta}=\cot\frac\theta2$$ That last equality comes from the half angle and double angle identities. Can you take it from here?

Answer 2

Noting that $1 + \cot^2(y) = \csc^2(y), $ you can convert your equation to $\sqrt{1 + \cot^2(y)} + \cot(y) = p$ and then $1 + \cot^2(y) = (p - \cot(y))^2,$ so $\cot(y) = \frac{p^2 - 1}{2p}.$

Answer 3

For finite non zero $p$,

$\csc y+\cot y=p$

$\iff\csc y-\cot y=1/p$

$\csc y=?,\cot y=?$

Alternatively, $\csc2A+\cot2A=\cdots=\cot A$